Answer
$\frac{14}{3}$
Work Step by Step
The volume under the surface $z=f(x,y)$ and above the region D in the xy-plane can be defined as:
$$Volume=\int\int_Df(x,y)dA$$
We can express the region D in the triangular region as follows:
$$D=\{(x,y)|0\leq x\leq2,0\leq y\leq2-x\}$$
Therefore,
$$Volume=\int\int_Df(x,y)dA
\\=\int_0^2\int_0^{2-x}(x^2+y^2+1)dydx
\\=\int_0^2\biggr[x^2y+\frac{1}{3}y^3+y\biggr]_{y=0}^{y=2-x}dx
\\=\int_0^2(-\frac{4}{3}x^3+4x^2-5x+\frac{14}{3})dx
\\=\biggr[-\frac{1}{3}x^4+\frac{4}{3}x^3-\frac{5}{2}x^2+\frac{14}{3}x\biggr]_0^2
\\=-\frac{16}{3}+\frac{32}{3}-10+\frac{28}{3}
\\=\frac{14}{3}$$
